What are the solutions to the equation (x-3)^2+(x+4)^2-(x-5)^2=17x+24 ?

The solutions to the equation (x-3)^2+(x+4)^2-(x-5)^2=17x+24 are x=8,x=-3

Find the zeros of (x-3)^2+(x+4)^2-(x-5)^2=17x+24

The zeros of (x-3)^2+(x+4)^2-(x-5)^2=17x+24 are x=8,x=-3

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(x - 3)^{2} + (x + 4)^{2} - (x - 5)^{2} = 17 x + 24

x = 8, x = - 3

Solution steps

(x - 3)^{2} + (x + 4)^{2} - (x - 5)^{2} = 17 x + 24

Expand

(x - 3)^{2} + (x + 4)^{2} - (x - 5)^{2} : x^{2} + 12 x

x^{2} + 12 x = 17 x + 24

Move

24

to the left side

x^{2} + 12 x - 24 = 17 x

Move

17 x

to the left side

x^{2} - 5 x - 24 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot ( - 24 )}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot (- 24) = 11

x_{1, 2} = \frac{- ( - 5 ) \pm 11}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 11}{2 \cdot 1}, x_{2} = \frac{- ( - 5 ) - 11}{2 \cdot 1}

x = \frac{- ( - 5 ) + 11}{2 \cdot 1} : 8

x = \frac{- ( - 5 ) - 11}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

x = 8, x = - 3

(4 - m)^{2} - 4 (3) (3 m - 36) = 0

x (2 x - 3) = 9

2 x^{2} - 12 x + 23 = 0

x^{2} + 16 x + 8 = 0

(x - 10)^{2} + 10 = 100

What are the solutions to the equation (x-3)^2+(x+4)^2-(x-5)^2=17x+24 ?
The solutions to the equation (x-3)^2+(x+4)^2-(x-5)^2=17x+24 are x=8,x=-3
Find the zeros of (x-3)^2+(x+4)^2-(x-5)^2=17x+24
The zeros of (x-3)^2+(x+4)^2-(x-5)^2=17x+24 are x=8,x=-3